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Question
Prove that: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
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Solution
\[\cos3x = 4 \cos^3 x - 3\text{ cos } x\]
\[ \Rightarrow \cos^3 x = \frac{\cos3x + 3\text{ cos } x}{4} . . . \left( i \right)\]
\[\text{ Also } , \]
\[\sin3x = 3\text{ sin } x - 4 \sin^3 x\]
\[ \Rightarrow \sin^3 x = \frac{3\text{ sin } x - \sin3x}{4} . . . \left( ii \right)\]
\[LHS = \cos^3 x \sin3x + \sin^3 x \cos3x\]
\[ = \left( \frac{\cos3x + 3\text{ cos } x}{4} \right)\sin3x + \left( \frac{3\text{ sin } x - \sin3x}{4} \right)\cos3x\]
\[ \left[ \text{ Using } \left( i \right) \text{ and } \left( ii \right) \right]\]
\[ = \frac{1}{4}\left[ 3\left( \sin3x \text{ cos } x + \text{ sin } x \cos3x \right) + \left( \cos3x \sin3x - \sin3x \cos3x \right) \right]\]
\[ = \frac{1}{4}\left[ 3\sin\left( 3x + x \right) + 0 \right]\]
\[ = \frac{3}{4}\sin4x\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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